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How does one prove that $n^2 +5n + 16$ is not divisible by $169$ for any integer $n$?

THOUGHTS:

This is equivalent to say that $$ n^2 +5n + 16=0\pmod{169} $$ has no solutions. One can also observe that $169=13^2$. And of course one cannot expect to prove this case by case since $\mathbb{Z}$ is not a finite set. But I really don't know how to proceed from here. Can any one help?

nonuser
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mhmhhmhmhm
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5 Answers5

9

This is one of my favorite elementary number theory problems.

Hint: $$f(n)=n^2+5n+16=(n^2+5n-36)+52$$

$f(n)=(n+9)(n-4)+52$. Assume that there is some $n$ such that $13\mid f(n)$. Then $13\mid n+9$ or $13\mid n-4$. But if one of those is true, then the other is as well, since $9\equiv-4\pmod{13}$. In other words, if $13\mid f(n)$ then $169\mid n^2+5n-36$. But if this is true, then $f(n)\equiv52\pmod{169}$.

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    I don't understand this hint – nonuser Oct 15 '19 at 19:25
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    @Aqua You can factor the quadratic expression. Then show that if the entire expression is divisible by 13, then it cannot be divisible by 169. –  Oct 15 '19 at 19:27
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    $(n+9)(n-4)+52$. So, $13\mid(n+9)(n-4)$, which means that $13|n+9$ or $13|n-4$. But if it divides either, it divides both, so $169\mid n^2+5n-36$, a contradiction. – Rushabh Mehta Oct 15 '19 at 19:27
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Suppose it is, then also $$13\mid n^2+5n+16$$ so $$13\mid (n^2+5n+16)-13n=n^2-8n+16$$

so $$ 13\mid n-4\implies 169\mid (n-4)^2 = n^2-8n+16$$

So $$169 \mid (n^2+5n+16)-(n^2-8n+16)= 13n\implies 13\mid n$$ But then from 1.st relation we get $$13\mid 16$$ a contradiction!

nonuser
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  • Wow, that's cute. – Rushabh Mehta Oct 15 '19 at 19:24
  • @Don Actually we can prove much more just as simply - see my answer. – Bill Dubuque Oct 16 '19 at 01:30
  • +1 because $n^2+5n+16=n^2-2\cdot4n+4^2+13n=(n-4)^2+13n$ is, thanks to $16=4^2$, surely easier to spot than $n^2+5n+16=n^2+5n-36+52=(n+9)(n-4)+52$. The question is bad, because getting the answer entails a trick that can't be derived, but can only be seen if you have the luck to make the correct mental leap. The easier the leap, the better. – Rosie F Apr 28 '21 at 13:20
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Let's try to complete the square. Equivalences below are $\bmod 169$.

$x^2+5x+16\equiv 0$

$4x^2+20x+64\equiv 0$

$4x^2+20x+25=(2x+5)^2\equiv 25-64=-39$

We need to find a quantity whose square is $\equiv -39$. Unfortunately this is a multiple of $13$ and the only square multiples of $13$ are also multiples of $169$ --therefore $\equiv 0\not\equiv -39$. And we're having a bad day.

Oscar Lanzi
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Let $n\in \Bbb Z$ be so that $n^2+5n+16$ is divisible by $13$. Then working modulo $13$ we have $$ \begin{aligned} n^2+5n+16 &\equiv n^2 + 18n + 81 \\ &= (n+9)^2 \qquad\text{ modulo }13\ . \end{aligned} $$ So $n$ is of the form $n=4+13k$, we substitute and get (computation in $\Bbb Z$): $$ \begin{aligned} n^2+5n+16 &= (13k +4)^2 + 5(13k+4) + 16 \\ &= 169k^2+169k+52\ . \end{aligned} $$ This is $52$ modulo $13^2$.

dan_fulea
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It's just as easy to prove a more general result (which is much more useful). The OP is special case

$\ \ a,b\,=\,n\!-\!4,\,n\ $ below $\,\Rightarrow \ p^2\mid n^2+(p\!-\!8)\,n+16\iff p = 2\mid n,\ $ so $\ p\neq 13\ \color{#c00}\checkmark $

Lemma $ $ If $\,p\,$ is $\rm\color{#c00}{prime}$ then $\,p^2\mid a^2\!+pb\iff p\mid a,b\,\smash[]{\overset{\ \rm\color{#0a0}U}\iff}\, p\mid (a,b) $

Proof $\,\ \ (\Leftarrow)\,\ $ Clear. $\,\ \ (\Rightarrow)\ $ $\ p^2\mid a^2\! + pb\,\Rightarrow\, p\mid a^2\color{#c00}{\Rightarrow}\, p\mid a\,$ $\Rightarrow\,p^2\mid a^2\Rightarrow p^2\mid pb\,\Rightarrow\,p\mid b$

The equivalence $\rm\color{#0a0}U$ is a special case of the GCD Universal Property.

Remark $ $ The Lemma is also true for for ${\rm\color{#c00}{squarefree}}\,p\,$ since they are precisely the integers satisfying the above middle inference: $\ p\mid a^2\Rightarrow\,p\mid a,\,$ for all integers $\,a.$

Bill Dubuque
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